# How do you solve abs(3x-4)<=x?

Jan 18, 2017

$| 3 x - 4 | \le x \implies x \in \left[1 , 2\right]$

or $1 \le x \le 2$

#### Explanation:

2 cases

$3 x - 4$ is positive

$| 3 x - 4 | \ge 0 \implies 3 x - 4 = 3 x - 4 \le x$

or

$3 x - 4$ is negative

$| 3 x - 4 | < 0 \implies | 3 x - 4 | = 4 - 3 x \le x$

If $3 x - 4$ is positive then,

$3 x - 4 \le x$

$\iff$

$3 x \le x + 4$ Add 4 to both sides

$\iff$

$3 x - x = 2 x \le 4$ Subtract x from both sides

$\iff$

$x \le 2$ Divide both sides by 2

Or $x \in \left(- \infty , 2\right]$

If $3 x - 4$ is negative then,

$4 - 3 x \le x$

$\iff$

$4 \le 4 x$ Add $3 x$ to both sides

$\iff$

$1 \le x$ Divide both sides by 4

Or $x \in \left[1 , \infty\right)$

Notice that

$\left(- \infty , 2\right] \cap \left[1 , \infty\right) = \left[1 , 2\right]$

Then

$x \in \left[1 , 2\right]$

Jan 18, 2017

$x \in \left[1 , 2\right]$

#### Explanation:

Here's one method...

Given:

$\left\mid 3 x - 4 \right\mid \le x$

Note in passing that $\left\mid \ldots \right\mid \ge 0$ and hence $x \ge 0$.

Given that $x \ge 0$, we can square both sides of the inequality to get:

$9 {x}^{2} - 24 x + 16 \le {x}^{2}$

Subtract ${x}^{2}$ from both sides to get:

$8 {x}^{2} - 24 x + 16 \le 0$

Divide both sides by $8$ to get:

${x}^{2} - 3 x + 2 \le 0$

$\left(x - 1\right) \left(x - 2\right) \le 0$

So this is a parabola with positive ${x}^{2}$ coefficient, intersecting the $x$ axis at $\left(1 , 0\right)$ and $\left(2 , 0\right)$

Hence the solution of our inequality is:

$1 \le x \le 2$

In interval notation:

$x \in \left[1 , 2\right]$

Jan 18, 2017

$x \in \left[1 , 2\right]$. See the segment of of the x-axisl in the Socratic graph. for this solution.

#### Explanation:

graph{(|3x-4|-x-y)<=0 [-5, 5, -2.5, 2.5]}

$| 3 x - 4 | \le x$ is the combined inequality for

$3 x - 4 \le x$, giving $x \le 2$, when $x \ge \frac{4}{3}$ and

$- \left(3 x - 4\right) \le x$, giving $x \ge 1$, when $x \le \frac{4}{3}$.